Question

If the point on the curve $y^2=6x,$ nearest to the point $(3,\frac{3}{2})$ is $(\alpha ,\beta )$ then $2(\alpha +\beta )$ is equal to

Answer 1

Let a point on $y^2=6x\text{is}P(\frac{3}{2}{t}^2,3t)$
The distance between $P$ and $(3,\frac{3}{2})\text{is D.}$

$\therefore D^2={(\frac{3t^2}{2}-3)}^2+{(3t-\frac{3}{2})}^2$

$=9(\frac{t^4}{4}-t^2+1+t^2-t+\frac{1}{4})$
$=\frac{9}{4}(t^4-4t+5)$
$\therefore 2D.\frac{dD}{dt}=\frac{9}{4}(4t^3-4)=9(t-1)(t^2+t+1)$
$\therefore \text{For}t=1,D^2\text{will be minimum.}$
$\therefore P=(\frac{3}{2},3)=(\alpha ,\beta )$
$\therefore 2(\alpha +\beta )=9$